Derivartive second question

[Ryan Rigdon]

y = ((x+9)^5) / (6x-3)

y ' = ?

could i rewrite it like this (x+9)^5 * 1/(6x-3)

then apply power and chain rule

 

[Deleted member 4993]

y = ((x+9)^5) / (6x-3)

y ' = ?

could i rewrite it like this (x+9)^5 * 1/(6x-3)

then apply power and chain rule

Sure you can do it that way....

However, I would use quotient rule , power rule and chain rule

\(\displaystyle \frac{d}{dx}\left [\frac{(x+9)^5}{6x-3}\right ]\ \ = \ \ \\\frac{5*(x+9)^4*(6x-3)\ \ - \ \ 6 * (x+9)^5}{(6x-3)^2}\ \ = \ \ \frac{(x+9)^4(8x-23)}{3(2x-1)^2}\)

 

[Ryan Rigdon]

my final answer was (5(x+9)^(4))/(3(2x-1)) - (2(x+9)^(5))/(3(2x-1)^(2))

 

[Deleted member 4993]

my final answer was (5(x+9)^(4))/(3(2x-1)) - (2(x+9)^(5))/(3(2x-1)^(2))... incorrect

If you do not factor out 3 from the denominator - prior to differentiating - you should have 6 in the numerator (as opposed to 2).

 

[Deleted member 4993]

[quote="Ryan Rigdon":3m1bdyzg]my final answer was (5(x+9)^(4))/(3(2x-1)) - (2(x+9)^(5))/(3(2x-1)^(2))... incorrect

If you do not factor out 3 from the denominator - prior to differentiating - you should have 6 in the numerator (as opposed to 2).[/quote:3m1bdyzg]

\(\displaystyle \frac{d}{dx}\left [\frac{(x+9)^5}{6x-3}\right ]\)

\(\displaystyle = \ \ \frac{d}{dx}\left [ (x+9)^5\cdot \frac{1}{6x-3}\right ]\)

\(\displaystyle = \ \ \frac{1}{3}\frac{d}{dx}\left [ (x+9)^5\cdot \frac{1}{2x-1}\right ]\)

\(\displaystyle = \ \ \frac{1}{3}\cdot \left [ 5\cdot (x+9)^4\cdot \frac{1}{2x-1} \ \ + \ \ (x+9)^5\cdot (-2)\cdot \frac{1}{(2x-1)^2}\right ]\)

and so on ....